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add Python solution
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@ -397,6 +397,9 @@ public:
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};
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```
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## Python
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# 105.从前序与中序遍历序列构造二叉树
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[力扣题目链接](https://leetcode.cn/problems/construct-binary-tree-from-preorder-and-inorder-traversal/)
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@ -650,6 +653,37 @@ class Solution {
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```
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## Python
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```python
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class Solution:
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def buildTree(self, inorder: List[int], postorder: List[int]) -> Optional[TreeNode]:
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# 第一步: 特殊情况讨论: 树为空. 或者说是递归终止条件
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if not postorder:
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return
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# 第二步: 后序遍历的最后一个就是当前的中间节点
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root_val = postorder[-1]
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root = TreeNode(root_val)
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# 第三步: 找切割点.
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root_index = inorder.index(root_val)
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# 第四步: 切割inorder数组. 得到inorder数组的左,右半边.
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left_inorder = inorder[:root_index]
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right_inorder = inorder[root_index + 1:]
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# 第五步: 切割postorder数组. 得到postorder数组的左,右半边.
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# ⭐️ 重点1: 中序数组大小一定跟后序数组大小是相同的.
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left_postorder = postorder[:len(left_inorder)]
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right_postorder = postorder[len(left_inorder): len(postorder) - 1]
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# 第六步: 递归
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root.left = self.buildTree(left_inorder, left_postorder)
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root.right = self.buildTree(right_inorder, right_postorder)
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# 第七步: 返回答案
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return root
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```
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105.从前序与中序遍历序列构造二叉树
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